# A BIPARTITE GRAPH ASSOCIATED WITH IRREDUCIBLE ELEMENTS AND GROUP OF UNITS IN \$\mathbb{Z}_n\$

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ISSN: 1311-1728 (printed version); ISSN: 1314-8060 (on-line version) doi:http://dx.doi.org/10.12732/ijam.v31i1.10

A BIPARTITE GRAPH ASSOCIATED WITH IRREDUCIBLE ELEMENTS AND GROUP OF UNITS IN Zn

Augustine Musukwa1§, Khumbo Kumwenda2 1Department of Mathematics

University of Trento

Via Della Malpensada 140, Trento TN, ITALY 1,2Department of Mathematics

Mzuzu University, MALAWI

Abstract: A nonzero nonunit aof a ring R is called an irreducible element if, for someb, c ∈R,a=bc implies that either b or c (not both) is a unit. We construct a bipartite graph in which the union of the set of irreducible elements and group of units is a vertex-set and an edge-set is the set of pairs between irreducible elements and their unit factors in the ring of integers modulo n. Many properties of this constructed bipartite graph are studied. We show that this bipartite graph contains components which are isomorphic. We also note that each component of this bipartite graph can be presented in some form which we call star form presentation. Some examples of graphs in star form presentation are provided for illustration purposes. Furthermore, we prove that the girth of this bipartite graph is 8. Most of the results in this paper are arrived at via group action.

AMS Subject Classification: 57M15, 68R10

Key Words: irreducible elements, units, bipartite graph, components, star form presentation

1. Introduction

In this paper we study a bipartite graph associated with irreducible elements

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and group of units in a ring of integers modulon. We construct a bipartite graph in which we define a vertex-set as the union of the set of irreducible elements and group of units and an edge-set as the set of pairs between irreducible elements and their unit factors. Our interest is to study and establish relationships between the set of irreducible elements and the group of units by using the properties of the graph we are going to construct. Since we are going to work with elements in the ring of integers modulon, in Section 2, we give an overview of this ring and we also give some results, available in [7], on how to determine the set of irreducible elements and its cardinality. We also present some results, which are crucial to our work, on group action and graph theory.

First part of Section 3 we are going to define some maps which are defined on group of units and show that the set of these maps is a group which acts on group of units. Our interest is to find orbits induced in group of units which are useful in proving some results in this paper. Lastly, we define and study a bipartite graph to an extent of determining some properties such as the components, degrees, connectivity, girth, circumference and others. More interestingly, we show that all components in this bipartite graph are isomorphic.

2. Preliminaries

In this section we give a quick overview of the ring of integers modulon, group action and graph theory, respectively.

2.1. An overview of the ring of integers modulo n

In this section the reader is referred to [3, 4, 5, 6, 7, 9] if more details are sought. We view the ring of integers modulo n, denotedZn, as the set {0,1, ..., n1} which is called the complete set of residues modulo n. Zn is a commutative ring with identity 1. For a nonzero Zn, we use the notation Z∗n. An element

u∈Zn is aunit if there exists vZn such thatuv = 1; in this casev is called a multiplicative inverse of u. All elements which are not units are said to be nonunits. The set of all units forms a group called the group of units. We denote a group of units by Un. Any element a is in the group of units of Zn if gcd(a, n) = 1, that is, the group of units is the set {a∈ Zn|gcd(a, n) = 1}. Euler’s phi function states that if n =Qk

i=1p

αi

i , a prime power factorization,

thenφ(n) =Qk

i=1(p

αi

i −p αi−1

i ) andφ(nm) =φ(n)φ(m) ifnand mare positive

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b, c∈R,a=bc implies that eitherbor c (not both) is a unit. In this paper we denote the set of irreducible elements byIn. Since in Zn elements do not have unique factors then a nonzero nonunit a is irreducible if all its factors satisfy the condition in the definition. For instance, 6 = 1·6 = 2·3 = 5·3 = 8·3 = 4·6 in Z9. So 6 is an irreducible element since all factors satisfy the condition in the definition.

Definition 1. Let n=Qk

i=1p

αi

i with pi distinct primes. We defineP as

the set of all pi withαi>1, i.e., P ={pi|pi is factor of nwithαi>1}.

To determine and count all irreducible elements in Zn we are going to use results in [7].

Theorem 2. Let n=Qk

i=1p

αi

i wherepi are distinct primes. Then

(i) In=∅, ifP =∅,

(ii)In={a∈Zn|gcd(a, n) =p, p∈ P},

(iii) |In|= P

p∈P

φ(n)

p .

Remark 3. Suppose that |P| 6= 0. Then, by Theorem 2 (ii), it is not hard to observe that the set of irreducible elements can as well be written as

In={up|u∈Un, p ∈ P}.

2.2. Group Action

Here we give an overview of group action which is a powerful tool in solving different problems in algebra and other branches of mathematics. Here we give some definitions and results which are relevant to what we study in this paper. If more details are sought, the reader is referred to [4, 6].

LetX be an arbitrary set, and letGbe a group. A functionf :G×X→X

is called group action by G on X if and only if ex = x for all x ∈ X and (g1g2)x =g1(g2x) for all g1, g2 ∈G and x∈X, wheree is the identity ofG. If

Gis acting on X then X is called a G-set. IfG acts on a set X and x, y∈X, thenx is said to be G-equivalent to y if there exists a g∈Gsuch thatgx=y. We write x ∼ y if two elements are G-equivalent. Let X be a G-set. Then

G-equivalence is an equivalence relation onX.

If X is a G-set, then each partition of X associated with G-equivalence is called anorbit of X under G. We denote the orbit that contains an elementx

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O={Ox|xX}. Let Gbe a group acting on a setX and letg be an element ofG. Then thefixed point set ofg inX, denoted byXg, is the set of allx∈X

such that gx= x. Note that Xg ⊆ X. The number of elements in the fixed point set of an element g ∈G is denoted by|Xg|and the number of orbits in

X is denoted by|O|. A group actsfaithfully on aG-set X if the identity is the only element of Gthat leaves every element of X fixed.

Theorem 4(Cauchy Frobenius Theorem). LetGbe a finite group acting

on a set X and let k denote the number of orbits in X under the action of G.

Then

k= 1

|G|

X

g∈G

|X(g)|.

2.3. Some Concepts on Graph Theory

In this section we consider some definitions and results in graph theory and the reader is referred to [1, 2, 8, 11] if more details are sought.

A simple graph G= (V, E) consists of a nonempty finite set V(G) of ele-ments calledverticesand a finite setE(G) of distinct unordered pairs of distinct elements ofV(G) callededges. We callV(G) thevertex-setandE(G) the edge-setof G. Each edge has a set of one or two vertices associated to it, which are called its endpoints and an edge is said to join its endpoints. Two edges of a graph are called adjacent if they share a vertex. Similarly, two vertices are called adjacent if they share an edge. An edge and a vertex on that edge are called incident. For a given vertex x, the number of all vertices adjacent to it is called degree of the vertexx, denoted byd(x). For a graphG, the minimum degree over all vertices is called the minimum degree of G, denoted by δ(G) and the maximum degree over all vertices is called the maximum degree of G, denoted by ∆(G).

If in a simple graph every pair of vertices are adjacent then the graph is called a complete graphand is denoted byKn. A graph with no edges is called an empty graph. If V′(G′) ⊆V(G) and E′(G′) ⊆E(G), then G′ = (V′, E′) is a subgraphof G. A graph is called connected if any two vertices are connected by some path; it is called disconnected otherwise. A connected subgraph H

is maximal provided H is not properly contained in a connected subgraph of

G. In other words, H is said to be a maximal connected subgraph of G if H

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by Cn where n is the number of vertices. If n is an even number then Cn is called even cycle. If nis odd then Cn isodd cycle. The girth of a graph is the length of a shortest cycle, denoted byg(G). Thecircumferenceof a graphGis the maximum length of a cycle inG, denoted by c(G)

Two graphsGandG′areisomorphicif there is a one-to-one correspondence between the vertices ofG, and those ofG′such that the number of edges joining any two vertices ofGis equal to the number of edges joining the corresponding vertices of G′. A graphG= (V, E) is calledbipartite if its vertex-set V(G) can be partitioned into two disjoint sets V1 and V2 in such a way that every edge connects vertices from different sets. We denoted it as G = (V1∪V2, E). A complete bipartite graphis a bipartite graph in which every vertex from partV1 is adjacent to every vertex from V2. If in a complete bipartite graph |V1|=r and |V2|=s, then the graph itself is denoted by Kr,s and the number of edges inKr,s equals rs. The complete bipartite graphK1,n is called a star.

3. Main Results

3.1. Group Action on group of units of Zn

In this section we are interested in determining and enumerating all orbits induced in Un under the action of a group which is shortly defined. Since in this paper we are concerned withZn which contains irreducible elements, from now on, we assume thatn=Qk

i=1p

αi

i with someαi >1 and distinct primespi,

i.e., we assume that |P| 6= 0. The results we obtain in this section are so useful in proving most of the results in the next section.

Definition 5. LetP be as defined in Definition 1. DefineA=Zp

1× · · · × Zp

|P| wherepi ∈ P.

Remark 6. It is well known that A is a group called direct product of groups under binary operation addition. It is not hard to observe that |A|=

Q

p∈P

p.

Definition 7. Let P be as defined in Definition 1. For u in Un and

s= (s1, ..., s|P|) inA, we define the mapsπsbyπs :u7→

P|P|

i=1sin/pi+uwhere

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Definition 8. We define the set of maps by ΘA={πs|s∈A}.

Note that when |P| = 1 then we simply haveA =Zp. In the next lemma we show that the set of maps ΘA is defined on group of unitsUn.

Lemma 9. πs is a map defined on Un.

Proof. We require that n = Qk

i=1p

αi

i with some αi > 1 and pi distinct

primes. Let u ∈ Un and suppose that πs(u) = ¯u such that ¯u 6∈ Un. This means that ¯umust be divisible by some pi. Without loss of generality, suppose it is divisible by p1, a factor of n. We can write ¯u = mp1 and πs(u)−u =

P|P|

i=1sin/pi =tp1 wheres= (s1, ..., s|P|) inA,pi∈ P, for some integers mand

t. Since P|P|

i=1sin/pi+u = ¯u implies u = ¯u−

P|P|

i=1sin/pi =p1(m−t) then

u /∈Un contradicting our earlier assumption that u ∈Un. This completes our

proof.

In the next lemma we show that the set of maps ΘA together with the

operation of composition of maps ◦form a group.

Lemma 10. ΘA together with the operation of composition of maps ◦is

a group.

Proof. Firstly, we show that ΘA is closed under the operation of

composition of maps◦. Letπs andπs′, with s= (s1, ..., s|P|) and s′ = (s

1, ..., s′|P|) in A, be any two elements of ΘA. So, foru∈Un,

πs◦πs′(u) =

|P|

X

i=1

sin/pi+

|P|

X

i=1

s′in/pi+u

=

|P|

X

i=1

(si+s′i)n/pi+u

= |P|

X

i=1

s′′in/pi+u=πs′′(u),

where s′′i = (si+si) (modpi). Thus it is closed under the operation of com-position of maps ◦.

Secondly, associativity follows from the associativity of mappings. Thirdly, it is clear that π(0,...,0) is the identity. Finally, given πs then it is not hard to see that its inverse π−1

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Theorem 11. ΘAacts onUn.

Proof. Letu∈Un. It is clear that π(0,...,0)(u) =u. Letπs and πs′, with s= (s1, ..., s|P|) and s′= (s′1, ..., s′|P|) in A, be any two elements of ΘA. Since

(πs◦πs′)(u) =πs′′(u) =

|P|

X

i=1

(si+s′i)n/pi+u

= |P|

X

i=1

sin/pi+

|P|

X

i=1

(s′i)n/pi+u

=πs(πs′(u)),

where s′′ = (s1+s′1, ..., s|P|+s′|P|), we conclude that ΘAacts on Un.

It worth noting that since it is only the identity in ΘA that leaves every

element in Un fixed, so ΘAacts faithfully on Un.

Definition 12. Letu∈Un. Then the orbit in Un containingu under the action of ΘA is Ou={πs(u)|s∈A}.

Lemma 13. For any u∈Un,|Ou|= Q p∈P

p.

Proof. Observe that, for u ∈Un, πs(u) =πs′(u) implies that s=s′. So it

should be easy to see that |Ou|must be equal to |A|. In Remark 6 we noted that|A|= Q

p∈P

p.

Let O denote the set of all orbits in Un under the action of ΘA, that is, O={Ou:uUn}.

Theorem 14. |O|= Qφ(n) p∈Pp.

Proof. We know thatO={Ou :uU}. All orbits in Un are of the same

size. Since|Un|=φ(n) and by Lemma 13,|Ou|= Q

p∈P

pso the result follows.

We consider two examples which illustrate the action of ΘA onUn.

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O4 = {4,9,14,19,24}. Observe that O = {O1,O2,O3,O4} and U25 = O1∪

O2∪ O3∪ O4.

Example 16. Consider Z36. Then we have ΘA = {πs|s A} where

A = Z2 ×Z3. So O1 = {1,7,13,19,25,31} and O5 = {5,11,17,23,29,35}. Observe thatO={O1,O5}and U36=O1∪ O5.

3.2. A Bipartite Graph Associated with In and Un

In this section we construct a bipartite graph which is associated with the set of irreducible elements and group of units in the ringZn. We begin by defining our new graph in line with the set of irreducible elements in Zn presented in Remark 3. Recall that the set of irreducible elements can be written as

In={up|u∈Un, p ∈ P}.

Definition 17. Let P be defined as in Definition 1. Define a graph

G= (V, E) as follows:

V(G) =In∪Un,

[v, w]∈E(G)⇔v∈Un and w=vp, p∈ P.

It is clear in our definition thatw∈ In. Recall that if|P|= 0, thenZndoes not contain irreducible elements (see Theorem 2) and so in such case we simply have an empty graph. We thus consider the ring which contains irreducible elements, i.e., |P| ≥ 1. Since this graph is associated with elements in Zn, we denote this graph by Gn instead of G, V(G) is denoted by Vn and E(G) is denoted by En.

Remark 18. Observe that since |In|= P

p∈P

φ(n)

p (by Theorem 2) and |Un|=φ(n) then |Vn|=φ(n)(P

p∈Pp−1+ 1).

Example 19. In Z8, we have U8 = {1,3,5,7}, I8 = {2,6} and E8 =

{[1,2],[3,6],[5,2],[7,6]}. SeeG8 in the figure that follows.

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2 6

1 5 3 7

Figure 1: A graph ofG8

2 3 10 14 15 21 22 26 33 34

1 5 7 11 13 17 19 23 25 29 31 35

Figure 2: A graph of G36

Example 21. In Z54, we have I54={3,15,21,33,39,51} and

U54={1,5,7,11,13,17,19,23,25,29,31,35,37,41,43,47,49,53}. We represent

G54 in the figure as follows.

3 15 21 33 39 51

1 19 37 5 23 41 7 25 43 11 29 47 13 31 49 17 35 53

Figure 3: A graph of G54

We next show how the set In is related to orbits induced in Un under the

action of ΘAand subsequently, use the established relationship in proving some

of the results which are of importance in this paper.

Lemma 22. Foru∈Un, if v∈ Ou is a unit factor ofw∈ In then all unit factors of w are in the sameOu.

Proof. A typical element inOu is of the formπs(u) =P|P|

i=1sin/pi+uwhere

s= (s1, ..., s|P|) in A, 0≤si ≤pi−1 and pi ∈ P (see Definition 7). Without loss of generality, suppose w is generated by p1, that is w = vp1 for v ∈ Ou. We need to show that all other unit factors of ware in Ou. Setv=v0 so that

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for all j (1 ≤ j ≤ p1−1), are also unit factors of w. Observe that for, j, k ∈

{0, ..., p1 −1}, vjp1 = π(j,s2,...,s|P|)(u)p1 = ( jn p1 +

s2n

p2 +· · ·+

s|P|n

p|P| +u)p1 =

(jn + s2n

p2 p1 +· · ·+

s|P|n

p|P| p1 +up1) ≡ (kn+ s2n

p2 p1 +· · ·+

s|P|n

p|P| p1 +up1) =

(kn p1 +

s2n

p2 +· · ·+

s|P|n

p|P| +u)p1 = π(k,s2,...,s|P|)(u)p1 = vkp1 (modn). That is w=vjp1, for all j (1 ≤j ≤p1−1). It must be not hard to see that these are the only factors inOu which are factors ofw.

We claim thatw does not contain any unit factor in other orbits. For sake of contradiction, supposewhas another unit factor v′

l=π(l,z2,...,z|P|)(u

) in Ou

where u′ ∈Un and u′ 6∈ Ou. That is vl′ ∈ Ou/ . It means that, for any vj ∈ Ou, a unit factor of w we have v′lp1 ≡ vjp1 (modn). This implies that v′l ≡ vj

(modn/p1) from which we get thatv′l=mn/p1+vj =mn/p1+π(j,s2,...,s|P|)(u) =

π(m+j,s2,...,s|P|)(u)∈ Ou contrary to our assumption that v

l6∈ Ou.

In the next lemma we show that if we let s1 = · · · = si−1 = si+1 =

· · · = s|P| = 0 and 0 ≤ si ≤ pi − 1 so that we have ΘA′ = {πs′|s′ =

(0, ...,0, si,0, ...,0), si ∈ A′}, where A′ = Zp

i and pi ∈ P, then further acting ΘA′ on Ou other orbits are induced. We will simply write ΘA′ ={πs′|s′ ∈A′}

whereA′ =Z

p. Foru′ ∈ Ou, we denote the orbit containingu′ under the action

of ΘA′ on Ou by Ouu′ = {sin/pi+u′|0 ≤ si ≤ p1 −1}. We also show that

elements in each Ouu′ are all unit factors of one irreducible element generated

by pi.

Lemma 23. Let ΘA′ ={πs

i|s ′

i ∈ A′} where A′ =Zpi and pi ∈ P. Then,

for u ∈ Un, ΘA′ acts on Ou and elements in each orbit are all unit factors of

one irreducible element generated by pi. Furthermore, these are the only unit

factors of this irreducible element.

Proof. It is not hard to see that ΘA′ is a subgroup of ΘA. Also note that

ΘA′ acts onOu since, for allu′ ∈ Ou,π0(u′) =u′ and πs(πs′(u′)) = (πsπs′)(u),

for all πs, πs′ ∈ ΘA′. Let the orbit containing u′ ∈ Ou be denoted by Ouu′ = {sin/pi+u′|0≤si ≤pi−1}. We show that if an irreducible element generated by pi has a factor in Ouu′ then all elements in Ouu′ are also factors of it.

Suppose thatv∈ Ouu′ such thatw=vpi. Writev=kn/pi+u′for somek∈ {0, ...pi−1}. For anyv′ inOuu′, we havev′ =gn/pi+u′whereg∈ {0, ..., pi−1}.

So v′pi = (gn/pi +u′)pi = gn+u′pi ≡ kn+u′pi = (kn/pi+u′)pi =vpi =w

(modn) implies that all elements in Ouu′ are factors ofw.

We claim thatOuu′ is the only orbit with factors ofw. For sake of

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ofw. So letOuu′′ ={sin/pi+u′′|0≤si ≤pi−1}and writev′′=tn/pi+u′′ for

somet∈ {0, .., pi−1}. Sov′′pi vpi =w (modn)(tn/pi+u′′)pi(kn/pi+

u′)pi (mod n)⇒ (tn/pi+u′′)pi ≡kn+u′pi (modn) ⇒(tn/pi+u′′)pi ≡u′pi

(modn) ⇒ tn/pi +u′′ ≡ u′ (modn/pi) ⇒ tn/pi +u′′ = hn/pi +u′. But

tn/pi+u′′ = hn/pi +u′ implies an element in Ouu′ is equal to an element in Ouu′′ which is a contradiction since Ouu′ and Ouu′′ two different orbits in Ou

under the action of ΘA′.

Remark 24. Suppose the orbit containing u′ ∈ Ou is Ouu′ ={sin/pi+ u′|0≤si≤pi−1} under the action of ΘA′, withA′=Zp

i. Then

(i) it is easily observed that |Ouu′|=pi from which we conclude that there

are

|Ou|/|Ouu′|=p1· · ·pi−1pi+1· · ·p|P| orbits inOu.

(ii) by (i) and Lemma 23, it implies that there are p1· · ·pi−1pi+1· · ·p|P| irre-ducible elements which are generated bypi and have their unit factors in Ou.

(iii) it follows from (ii) that if |P| > 1 then there are P|P|

i=1σi, where σi =

p1· · ·pi−1pi+1· · ·p|P|, irreducible elements which have factors in Ou and 1 if|P|= 1.

Lemma 25. LetA′=Z

pi andA ′′ =Z

pj, fori6=j. Then all orbits induced

by ΘA′ and ΘA′′ inOu where u∈Un contain at most 1 element in common.

Proof. Suppose some orbits induced inOu by the actions of ΘA′ and ΘA′′

contain more than one element in common. Supposeu′ is one of the elements in such orbits. So we have{sin/pi+u′|0sipi1}as an orbit induced by Θ

A′

and {sjn/pj+u′|0≤sj ≤pj−1} as an orbit induced by ΘA′′. We show that u′ must be the only element in common. Suppose sin/pi +u′ = sjn/pj+u′, for some si ∈ {0, ..., pi−1}and sj ∈ {0, ..., pj−1}. Ifsi =sj = 0 we obviously have u′ which we already know that is present in both orbits. Supposesi and

sj are both nonzero. Since sin/pi+u′ =sjn/pj+u′ implies sipj =sjpi then

pj divides either sj or pi and pi divides eithersi or pj which is a contradiction aspi 6=pj,si< pi and sj < pj.

Theorem 26. Gncontains Qφ(n)

p∈Pp isomorphic components.

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element has a unit factor in Ou where u ∈ Un then all factors for that irre-ducible element are also in the same orbit. We are assured that the component associated withOu is connected because all elements in an orbit induced under the action of ΘA′ on Ou occur as elements in other orbits induced under the

actions of ΘA′′ (we know that A′ = Zp

i and A ′ = Z

pi with i6= j, see Lemma 25).

Since all orbitsOu inUn when acted upon by ΘAhave the same structure

then all components obtained from these orbits must be isomorphic. Since, by Lemma 13, |Ou| = Q

p∈Pp then we conclude that there are φ(n)/

Q

p∈Pp components.

From Theorem 26 we remark that the graph Gn is a disconnected graph with isomorphic components. Since all the components in Gn are isomorphic

then the whole graph can be described by a single component. For this reason, in the rest of this section we mostly use a component to study properties ofGn.

Lemma 27. No two irreducible elements in In generated by the same

prime are adjacent to a common unit in Un.

Proof. This is a direct consequence of Lemma 23. Also by Remark 3, any

irreducible element generated by p can be written in the form up where u ∈ Un and so with this presentation it must be impossible for any two elements

generated pto be adjacent to a common unit in Un.

Proposition 28. InGn,d(v) =|P|, for allv∈Unandd(w) =pifw=v′p

for somev′∈Un and p∈ P.

Proof. By lemma 27, no two irreducible element generated by the same

prime are adjacent to a common unit factor. But irreducible elements generated by different primes can have common unit factor. So each unit in Un must be adjacent to|P| irreducible element.

By Remark 24, if the orbit containing u′ ∈ Ou is Ouu′ ={sin/pi+u′|0≤ si ≤pi−1} under the action of ΘA′ then we have |Ouu′|=pi. By Lemma 23,

we conclude that only elements in Ouu′ are adjacent to one irreducible element

generated bypi. So if such irreducible element isw then thed(w) =pi.

Remark 29. Since, by Proposition 28, any vertex in In has degree p

where p ∈ P and any vertex in Un has degree |P| then δ(Gn) = min

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and ∆(Gn) = max

p∈P{p,|P|}.

Proposition 30. The total sum of degrees for all vertices in Gn is

2φ(n)|P|.

Proof. Since, by Proposition 28, the degree for each vertex in Un is |P|

and there are φ(n) elements in Un then the sum of degrees on unit vertices is

φ(n)|P|. Again all irreducible elements generated by prime p have degree p. Since there are φ(pn) irreducible elements generated by p then they contribute

φ(n) degrees. Since there are |P| primes which generate In then the sum of degrees on irreducible vertices is φ(n)|P|. Thus Gn has a total of 2φ(n)|P|

degrees.

Corollary 31. The total number of edges in Gn is φ(n)|P|.

Proof. There areφ(n) elements inUn. By Lemma 28, each unit vertex has

degree |P|. HenceGn hasφ(n)|P| edges.

Proposition 32. A component of Gn is a union of copies of the star K1,|P|.

Proof. By Proposition 28, the degree of any unit inUn is|P|so the results

follows.

Remark 33. By Proposition 28, the degree of an irreducible element generated by p is p so Proposition 32 is also equivalent to saying that each component is a union of copies of the starsK1,p wherep∈ P.

Proposition 34. If P ={p}then each component of Gn isK1,p.

Proof. SupposeP ={p}. Then all irreducible elements inInare generated

by the same primep. Hence by Lemma 27 and Proposition 28, each irreducible element is adjacent to p units and these units are not adjacent to any other irreducible element. Thus the result follows.

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Proof. Suppose that Gn contains a cycle. It implies that some vertices in both In and Un have at least degree 2. |P|= 0 implies that Gn is an empty graph. If|P|= 1,Gndoes not contain a cycle since in this case all unit vertices have degree 1. Since if |P|>1 all vertices in both In and Un contain at least degree 2 then Gn must contain a cycle. The converse is obvious.

Theorem 36. Let P = {p1, p2}. Then between any two irreducible

elements generated by prime p ∈ P there are p paths of length 4 in each

component of Gn. Furthermore, there are p(p2−1) cycles of length 8 that pass through any such two irreducible elements.

Proof. Suppose P = {p1, p2}. By Theorem 2, In is generated by p1 and

p2. Since |P| = 2 >1 then, by Theorem 35, Gn contains cycles. By Remark 3, recall that an irreducible element generated by p can be written as up, for someu ∈ Un. Consider the maps π(s,t) with (s, t) ∈A =Zp1 ×Zp2 as defined in Definition 7. Denote by ws all irreducible elements generated by p2 and

w′

t those generated by p1 where 0 ≤ s ≤ p1 −1 and 0 ≤ t ≤ p2 −1. For

i 6= j, π(i,t)(u)p1 = (inp1 + tnp2 +u)p1 = in+ p1ptn2 +up1 ≡ jn+ pp1tn2 +up1 = (jnp

1 +

tn

p2 +u)p1 =π(j,t)(u)p1 (modn) implies that if w ′

t is adjacent toπ(i,t)(u) then it is also adjacent toπ(j,t)(u). Similarly, forg6=f, ifwsis adjacentπ(s,g)(u) then it is also adjacent to π(s,f)(u). So it is immediate that between any two irreducible elements generated by p2 (this is also true with those generated by

p1) we have the paths below:

wi, π(i,t)(u), wt′, π(j,t)(u), wj

for all 0≤i < j ≤p1−1 and 0≤t≤p2−1. We represent these paths in Figure 4 below.

It is not hard to observe, in Figure 4, that between any two irreducible elements generated by p2 there are p2 paths of length 4, each passing through some w′

t where 0 ≤ t ≤ p2 −1 and similarly, between any two irreducible elements generated by p1 there are p1 paths of length 4. Thus it is immediate that any two of the pj paths of length 4 between any two irreducible elements generated bypj form a cycle of length 8 which pass through such two irreducible

elements. So there are pj 2

= pj(p2j−1) such cycles.

Remark 37. Letn=Qk

i=1p

αi

i with some αi >1 and pi distinct primes.

If you consider the maps π(s1,...,s|P|) : u 7→

P|P|

j=1

sjn

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wi wj w′0

w′1

w′ 2 .. .

w′ (p2−1)

π(i,0)(u)

π(i,1)(u)

π(i,2)(u)

.. .

π(i,p2−1)(u)

π(j,0)(u)

π(j,1)(u)

π(j,2)(u)

.. .

π(j,p2−1)(u)

Figure 4: Paths between wi and wj in a component of Gn

we haveπ(si,sj) then arguing by Theorem 36 one can easily find two irreducible elements generated by same prime pj in a component of Gn with pj paths of length 4 between them and pj(p2j−1) cycles of length 8 that pass through them.

Theorem 36 and Remark 37 suggest that any component of Gn can be presented in some form. From Figure 4 we observe that any component of Gn

can be presented in a form which we will callstar form presentationby putting all elements generated by any prime in the center and all others around them. If the irreducible elements put in the center are those that are generated bypthen our graph in star form presentation takes the shape of K1,p since the degree of

each such irreducible elements is p by Proposition 28. To illustrate this idea, we provide three examples and in each we present a component ofGn in a star

form presentation. Since if|P|= 1, by Proposition 34, each component is a star then we already have a star form presentation. This can also be observed in Examples 19 and 21. We therefore consider examples in which we have|P|>1. In the examples to be given, the blue vertices are for irreducible elements and the red vertices are for units. Each irreducible element can be identified by its degree since by Proposition 28 we know that any irreducible element generated by phas a degree p.

Example 38. Let n = 3α5β or n= 3α5βQk

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orbit under the action of ΘA′′ withA′′=Z5.

Figure 5: A component of Gn

irreducible elements generated by 5 in center

Figure 6: A component of Gn

with irreducible elements gener-ated by 3 in center

Example 39. Let n = 5α7β or n = 5α7βQk

i=1pi with α, β > 1, that is, P = {5,7}. Gn contains φ35(n) isomorphic components. Figure 7 shows a

component in a star form presentation with irreducible elements generated by 5 put in the center. Note that in Figure 7 all 7 units in each “branch” form some orbit in Ou under the action of ΘA′ with A′ = Z7. We could also put

irreducible elements generated by 7 in the center; in this case 7 “branches” are expected and in each “branch” we expect 5 units from some orbit in Ou under the action of ΘA′′ withA′′=Z5.

Figure 7: A component of Gn in star form presentation with irre-ducible elements generated by 5 in center

Example 40. We consider a component of Gn when n= 3λ5α7β or n= 3λ5α7βQk

i=1piwithλ, α, β >1 and there are

φ(n)

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irreducible elements generated by 5 or 7 in the center.

Figure 8: A component of Gn with irreducible elements generated by 3 put in the center

We point out few things in the construction of Figure 8. Red vertices are units all fromOu. All unit vertices in each column form an orbit in Ou under the action of ΘA′ where A′ =Z5 and in this case there 21 = 3×7 orbits. Also

all unit vertices in each row form an orbit inOu under the action of ΘA′′where A′′=Z

7 and there are 15 = 3×5 orbits. Both cases satisfy Remark 24 (i). As it was observed in Lemma 25, orbits induced by ΘA′ and ΘA′′ have at

most one element in common and this element is in the intersection of a row and a column. And the star form presentation is always possible because of the just said fact about orbits in each Ou . For |P|>3, the star form presentation is more funnier and complex. For instance, if|P|= 4 then unit vertices in each “branch” have to be arranged in 3 dimensional.

Theorem 41. If|P|>1then g(Gn) = 8.

Proof. Suppose that |P| > 1. So by Proposition 35 we are assured that

there are cycles. In bipartite graphs cycles always have even length so we expect an even girth. It is clear from Definition 17 thatGnis a simple graph so we cannot have cycles of length 2. We claim that cycles of length 4 and 6 are not possible. Suppose we have a cycle of length 4 as in Figure 9 where r1, r2 are two different irreducible elements andu1, u2 are two different unit vertices. Observe that r1 and r2 are not generated by the same prime since that would contradict Lemma 27. Without loss of generality, assume that r1 and

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r1 r2

u1 u2

Figure 9: A cycle of length 4

w1 w2 w3

v1 v2 v3

Figure 10: A cycle of length 6

(modn) which imply that u1 ≡u2 (mod n/p1) and u1 ≡u2 (modn/p2) from which obtainu1=u2+ns1/p1 andu1 =u2+ns2/p2for somes1∈ {0, ..., p1−1} and s2 ∈ {0, ..., p2 −1}. It follows that u1 =u2+ns1/p1 =u2+ns2/p2 from which we obtain s1p2 = s2p1. So it implies that p1 divides s1 or p2 and p2 divides s2 or p1 which is only possible if s1 = s2 = 0, meaning that u1 = u2 contrary to our assumption that they are different.

Next we suppose that we have a cycle of length 6 as in Figure 10 where

w1, w2, w3 are three different irreducible elements and v1, v2, v3 are three dif-ferent unit vertices. All the three irreducible elements must be generated by different primes otherwise it would mean that some irreducible elements gen-erated by the same prime have a common unit factor contrary to Lemma 27. So we further suppose that|P|>2 otherwise there is nothing to prove. With-out loss of generality, assume that w1, w2 and w3 are generated by p1, p2 and

p3, respectively. That is w1 = v1p1 ≡ v2p1 (mod n), w2 = v1p2 ≡ v3p2 (modn) and w3 = v2p1 ≡ v3p1 (modn). We note that w1 = v1p1 ≡ v2p1 (modn) ⇒ v1 ≡v2 (modn/p1) ⇒ v1 = v2+ns1/p1 where 0 ≤s1 ≤ p1 −1. Similarly, w2 = v1p2 ≡ v3p2 (modn) implies that v1 = v3 +ns2/p2 where 0≤s2 ≤p2−1 and w3 =v2p1 ≡v3p1 (modn) implies that v2 =v3+ns3/p2 where 0≤s3 ≤p3−1. So we have the equations:

v1 = v2 + ns1/p1 · · · (i)

v1 = v3 + ns2/p2 · · · (ii)

v2 = v3 + ns3/p3 · · · (iii)

By (i) and (ii) we obtain:

v2+ns1/p1 = v3+ns2/p2 · · · (iv)

Substituting (iii) in (iv) we obtain:

v3+ns3/p3 +ns1/p1 = v3+ns2/p2 · · · (v)

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s3p1p2+s1p2p3 = s2p1p3 · · · (vi)

It follows that p1 must divide left side of Equation (vi). But p1|(s3p1p2+

s1p2p3)⇒p1|s1p2p3. Since 0≤s1≤p1−1 so the only possibility is thats1 = 0 and from Equation (i) it implies thatv1=v2 contrary to our assumptionv1, v2 and v3 are different. Thus a cycle of length 6 is also impossible.

By Theorem 36 and Remark 37, if |P| > 1 there always exist a cycle of length 8 which passes through some two irreducible elements generated by the same prime. Hence we conclude thatg(Gn) = 8.

In Examples 38, 39 and 40 one can easily check thatg(Gn) = 8.

In the next proposition we discuss about the circumference of Gn when

|P|= 2.

Proposition 42. Let P ={p1, p2}. Ifp1 < p2 thenc(Gn) = 4p1.

Proof. By Remark 24 (ii), we conclude that in each component there are

p1 irreducible elements generated by p2 and p2 irreducible elements generated by p1. Since p1 < p2 then a possible longest cycle must contain all the p1 irreducible elements generated byp2and also p1 irreducible elements generated byp1 otherwise if a cycle contains more thanp1 irreducible elements generated by p1 it would mean that some two irreducible elements generated by p1 have a common unit factor contrary to Lemma 27. Denote by ws all irreducible elements generated byp2 and byw′tthose generated byp1 where 0≤s≤p1−1 and 0≤t≤p2−1. As in Theorem 36 , fori6=j,π(i,t)(u)p1 = (inp1+

tn

p2+u)p1=

in+p1tn

p2 +up1 ≡jn+

p1tn

p2 +up1= (

jn p1+

tn

p2+u)p1=π(j,t)(u)p1 (mod n) implies that if w′t is adjacent toπ(i,t)(u) then it is also adjacent to π(j,t)(u). Similarly, forl 6=k, ifws is adjacent π(s,l)(u) then it is also adjacent to π(s,k)(u). Hence the cycle that follows is always possible:

w0, π(0,0)(u), w0′, π(1,0)(u), w1, π(1,1)(u), w′2, π(2,1)(u),

w2, π(2,2)(u), w2′, π(3,2)(u), w3, π(3,3)(u), w′3, π(4,3)(u), (∗)

w4,· · · , w(p1−1), π(p1−1,p1−1)(u), w ′

(p1−1), π(p−1,0)(u), w0.

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of irreducible elements generated by p1 and p2. If there exist a cycle which is longer than the cycle in (∗) then it means that it includes more irreducible elements generated by p1 and that would mean some two irreducible elements generated by p1 are adjacent to one unit contrary to Lemma (27). So the cycle in (∗) must be a longest cycle in Gn.

Observe that between any two consecutive irreducible elements generated by the same prime in the cycle in (∗) there there is a path of length 4 so its clear that the length of this cycle must be 4p1. So we conclude that the circumference of Gnis 4p1.

It can easily be checked thatc(Gn) of graphs in Examples 38 and 39 satisfy Proposition 42.

Lastly, we provide a conjecture about the circumference of Gn which was arrived at by using an idea of star form presentation and also by using some insights from Theorem 36, Remarks 24 (ii) and 37 and Proposition 42.

Conjecture 43. Let |P|>1. Then c(Gn) = 4

Q

p∈Pp

max(P) .

4. Conclusion

In this paper we constructed a bipartite graph with the union of the set of irreducible elements and group of units as a vertex-set and the set of pairs between irreducible elements and their unit factors as an edge-set in the ring of integers modulo n. Many properties of this graph were studied. We proved that this bipartite graph is not connected but contains isomorphic components, each of which is a union of copies of the star K1,|P| where P is a set which contains all prime factors of n whose powers are greater than 1. For a given value of n, we determined when this graph has cycles or not. Furthermore, we proved that the girth of this bipartite graph is 8.

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